Problem: There are some tasks which have the following structure: you are given a model, and you can do some operations, you should use these operations to achive the goal. One way to create a new task is to use the same model and same operations, but change the goal.

Let's have a try. I have created the following task for Topcoder SRM 557 Div1-Hard: you are given n integers x1, x2, ..., xn. You are allowed to perform the assignments (as many as you want) of the following form xi ^= xj (in the original task i and j must be different, but in this task we allow i to equal j). The goal is to maximize the sum of all xi.

Now we just change the goal. You are also given n integers y1, y2, ..., yn. You should make x1, x2, ..., xn exactly equal to y1, y2, ..., yn. In other words, for each i number xi should be equal to yi.

Input

The first line contains an integer n (1 ≤ n ≤ 10000). The second line contains n integers: x1 to xn (0 ≤ xi ≤ 109). The third line contains n integers: y1 to yn (0 ≤ yi ≤ 109).

Output

If there is no solution, output -1.

If there is a solution, then in the first line output an integer m (0 ≤ m ≤ 1000000) – the number of assignments you need to perform. Then print m lines, each line should contain two integers i and j (1 ≤ i, j ≤ n), which denote assignment xi ^= xj.

If there are multiple solutions you can print any of them. We can prove that under these constraints if there exists a solution then there always exists a solution with no more than 106 operations.

Examples

Input

2
3 5
6 0


Output

2
1 2
2 2


Input

5
0 0 0 0 0
1 2 3 4 5


Output

-1


Input

3
4 5 6
1 2 3


Output

5
3 1
1 2
2 2
2 3
3 1


Input

3
1 2 3
4 5 6


Output

-1

Note

Assignment a ^= b denotes assignment a = a ^ b, where operation "^" is bitwise XOR of two integers.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int n;
vector<pair<int, int> > res1, res2, intermed;
int v[10004], v1[10004];
void swapa(int &a, int &b, int i, int j, vector<pair<int, int> > &res) {
  a ^= b;
  b ^= a;
  a ^= b;
  res.push_back(make_pair(i, j));
  res.push_back(make_pair(j, i));
  res.push_back(make_pair(i, j));
}
bool exist(int a, int bt) {
  if (a & (1 << (30 - bt))) return true;
  return false;
}
void gauss(int *v, vector<pair<int, int> > &res) {
  int onde = 0;
  for (int i = 0; i < 31; i++) {
    for (int j = onde; j < n; j++) {
      if (exist(v[j], i)) {
        if (onde != j) swapa(v[onde], v[j], onde, j, res);
        for (int k = j + 1; k < n; k++) {
          if (exist(v[k], i)) {
            v[k] ^= v[onde];
            res.push_back(make_pair(k, onde));
          }
        }
        onde++;
        break;
      }
    }
  }
}
void print(vector<pair<int, int> > &ou) {
  for (int i = 0; i < ou.size(); i++) {
    cout << ou[i].first + 1 << " " << ou[i].second + 1 << "\n";
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> v[i];
  }
  for (int i = 0; i < n; i++) {
    cin >> v1[i];
  }
  gauss(v, res1);
  gauss(v1, res2);
  reverse(res2.begin(), res2.end());
  for (int i = 0; i < min(40, n); i++) {
    int atual = v[i] ^ v1[i];
    for (int j = 0; j < min(40, n); j++) {
      if ((atual ^ v[j]) < atual) {
        atual ^= v[j];
        intermed.push_back(make_pair(i, j));
      }
    }
    if (atual != 0) {
      cout << -1 << "\n";
      return 0;
    }
  }
  cout << (long long)res1.size() + (long long)res2.size() +
              (long long)intermed.size()
       << "\n";
  print(res1);
  print(intermed);
  print(res2);
}